Subject: Chemistry — Stoichiometry
Question: In a one-litre flask, 6 moles of A under goes the
reaction $A(g) \rightleftharpoons P(g)$. The progress of product
formation at two temperatures (in Kelvin), $T_{1}$ and $T_{2}$, is shown
in the figure:
[IMAGE] If $T_{1} = 2T_{2}$ and
$\left( \mathrm{\Delta}G_{2}^{\ominus} - \mathrm{\Delta}G_{1}^{\ominus} \right) = RT_{2}\ln x$,
then the value of $x$ is \_\_\_\_\_\_.
\[$\mathrm{\Delta}G_{1}^{\ominus}$ and
$\mathrm{\Delta}G_{2}^{\ominus}$ are standard Gibb's free energy
change for the reaction at temperatures $T_{1}$ and $T_{2}$,
respectively.\]
Answer: (8)
Solution: $\mathrm{\Delta}G_{1}^{\ominus} = - RT_{1}\ln\frac{4}{2}$
$$\mathrm{\Delta}G_{2}^{\ominus} = - RT_{2}\ln\frac{2}{4}$$
$$\mathrm{\Delta}G_{2}^{\ominus} - \mathrm{\Delta}G_{1}^{\ominus}$$
$$= - RT_{2}\ln{\left( \frac{1}{2} \right) + RT_{1}\ln 2}$$
$$= - RT_{2}\ln{\left( \frac{1}{2} \right) + R_{2}T_{2}\ln 2}$$
$$= RT_{2}\ln\frac{4}{1/2}$$
$$RT_{2}\ln 8 = RT_{2}\ln x$$
⇒ $x = 8$